Exam (elaborations) TEST BANK FOR Quantum Physics 3rd Edition By Stephen Gasiorowicz (Solution manual)

Exam (elaborations) TEST BANK FOR Quantum Physics 3rd Edition By Stephen Gasiorowicz (Solution manual) SOLUTIONS MANUAL CHAPTER 1 1. The energy contained in a volume dV is U(ν,T)dV = U(ν ,T)r2drsinθ dθdϕ when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area dA is dE(ν ,T) = U(ν ,T)dV dAcosθ 4πr2 The total energy emitted is . dE(ν ,T) = dr dθ dϕU(ν,T)sinθ cosθ dA 0 4π 2π ∫ 0 π /2 ∫ 0 cΔt ∫ = dA 4π 2π cΔtU(ν ,T) dθ sinθ cosθ 0 π / 2 ∫ = 1 4 cΔtdAU (ν ,T) By definition of the emissivity, this is equal to EΔtdA . Hence E(ν,T) = c 4 U(ν,T) 2. We have w(λ ,T) = U(ν ,T) | dν / dλ |= U( c λ ) c λ2 = 8πhc λ 5 1 ehc/λkT −1 This density will be maximal when dw(λ ,T) / dλ = 0. What we need is d dλ 1 λ 5 1 eA /λ −1 ⎛ ⎝ ⎞ ⎠ = (−5 1 λ6 − 1 λ5 eA /λ eA/λ −1 (− A λ 2 )) 1 eA /λ −1 = 0 Where A = hc / kT . The above implies that with x = A /λ , we must have 5 − x = 5e−x A solution of this is x = 4.965 so that λ maxT = hc 4.965k = 2.898 ×10−3m In example 1.1 we were given an estimate of the sun’s surface temperature as 6000 K. From this we get λ max sun = 28.98 ×10−4mK 6 ×103K = 4.83 ×10−7m = 483nm 3. The relationship is hν = K +W where K is the electron kinetic energy and W is the work function. Here hν = hc λ = (6.626 ×10−34 J .s)(3×108m / s) 350 ×10−9m = 5.68 ×10−19J = 3.55eV With K = 1.60 eV, we get W = 1.95 eV 4. We use hc λ1 − hc λ2 = K1 − K2 since W cancels. From ;this we get h = 1 c λ1 λ2 λ2 − λ 1 (K1 − K2) = = (200 ×10−9m)(258 ×10−9m) (3×108m / s)(58 ×10−9m) × (2.3− 0.9)eV × (1.60 ×10−19)J / eV = 6.64 ×10−34 J .s 5. The maximum energy loss for the photon occurs in a head-on collision, with the photon scattered backwards. Let the incident photon energy be hν , and the backwardscattered photon energy be hν' . Let the energy of the recoiling proton be E. Then its recoil momentum is obtained from E = p2c 2 + m2c 4 . The energy conservation equation reads hν + mc2 = hν '+E and the momentum conservation equation reads hν c = − hν ' c + p that is hν = −hν '+ pc We get E + pc − mc2 = 2hν from which it follows that p2c2 + m2c4 = (2hν − pc + mc2)2 so that pc = 4h2ν 2 + 4hνmc2 4hν + 2mc2 The energy loss for the photon is the kinetic energy of the proton K = E − mc2 . Now hν = 100 MeV and mc2 = 938 MeV, so that pc = 182MeV and E − mc2 = K = 17.6MeV 6. Let hν be the incident photon energy, hν' the final photon energy and p the outgoing electron momentum. Energy conservation reads hν + mc2 = hν '+ p2c2 + m2c4 We write the equation for momentum conservation, assuming that the initial photon moves in the x –direction and the final photon in the y-direction. When multiplied by c it read i(hν ) = j(hν ') + (ipxc + jpyc) Hence pxc = hν ; pyc = −hν '. We use this to rewrite the energy conservation equation as follows: (hν + mc 2 − hν ')2 = m2c 4 + c 2(px 2 + py 2) = m2c4 + (hν )2 + (hν ')2 From this we get hν'= hν mc2 hν + mc2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ We may use this to calculate the kinetic energy of the electron K = hν − hν '= hν 1− mc2 hν + mc2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = hν hν hν + mc2 = (100keV )2 100keV + 510keV = 16.4keV Also pc = i(100keV ) + j(−83.6keV) which gives the direction of the recoiling electron. 7. The photon energy is hν = hc λ = (6.63×10−34 J.s)(3 ×108m / s) 3×106 ×10−9m = 6.63×10−17J = 6.63×10−17 J 1.60 ×10−19 J / eV = 4.14 ×10−4 MeV The momentum conservation for collinear motion (the collision is head on for maximum energy loss), when squared, reads hν c ⎛ ⎝ ⎞ ⎠ 2 + p2 + 2 hν c ⎛ ⎝ ⎞ ⎠ p η i = hν ' c ⎛ ⎝ ⎞ ⎠ 2 + p'2 +2 hν ' c ⎛ ⎝ ⎞ ⎠ p' η f Here η i = ±1, with the upper sign corresponding to the photon and the electron moving in the same/opposite direction, and similarly for η f . When this is multiplied by c2 we get (hν )2 + (pc)2 + 2(hν ) pc η i = (hν ')2 + ( p'c)2 + 2(hν ') p'c η f The square of the energy conservation equation, with E expressed in terms of momentum and mass reads (hν )2 + (pc)2 + m2c 4 + 2Ehν = (hν ')2 + ( p'c)2 + m2c4 + 2E' hν ' After we cancel the mass terms and subtracting, we get hν(E − η i pc) = hν '(E'− η f p'c) From this can calculate hν' and rewrite the energy conservation law in the form E − E'= hν E − η i pc E'−p'c η f −1 ⎛ ⎜⎝ ⎞ ⎟⎠ The energy loss is largest if η i = −1; η f = 1. Assuming that the final electron momentum is not very close to zero, we can write E + pc = 2E and E'− p'c = (mc2 )2 2E' so that E − E'= hν 2E × 2E' (mc2 )2 ⎛ ⎜⎝ ⎞ ⎟⎠ It follows that 1 E' = 1 E +16hν with everything expressed in MeV. This leads to E’ =(100/1.64)=61 MeV and the energy loss is 39MeV. 8.We have λ’ = 0.035 x 10-10 m, to be inserted into λ '−λ = h mec (1− cos600) = h 2mec = 6.63 ×10−34 J.s 2 × (0.9 ×10−30kg)(3×108m / s) = 1.23×10−12m Therefore λ = λ’ = (3.50-1.23) x 10-12 m = 2.3 x 10-12 m. The energy of the X-ray photon is therefore hν = hc λ = (6.63×10−34 J .s)(3 ×108m / s) (2.3×10−12m)(1.6 ×10−19 J / eV) = 5.4 ×105eV 9. With the nucleus initially at rest, the recoil momentum of the nucleus must be equal and opposite to that of the emitted photon. We therefore have its magnitude given by p = hν / c , where hν = 6.2 MeV . The recoil energy is E = p2 2M = hν hν 2Mc2 = (6.2MeV) 6.2MeV 2 ×14 × (940MeV ) = 1.5 ×10−3MeV 10. The formula λ = 2asinθ / n implies that λ / sinθ ≤ 2a / 3. Since λ = h/p this leads to p ≥ 3h / 2asinθ , which implies that the kinetic energy obeys K = p2 2m ≥ 9h2 8ma2 sin2θ Thus the minimum energy for electrons is K = 9(6.63×10−34 J.s)2 8(0.9 ×10−30 kg)(0.32 ×10−9m)2 (1.6 ×10−19 J / eV) = 3.35eV For Helium atoms the mass is 4(1.67 ×10−27 kg) / (0.9×10−30kg) = 7.42 ×103 larger, so that K = 33.5eV 7.42 ×103 = 4.5 ×10−3eV 11. We use K = p2 2m = h2 2mλ2 with λ = 15 x 10-9 m to get K = (6.63×10−34 J.s)2 2(0.9 ×10−30 kg)(15 ×10−9m)2 (1.6 ×10−19 J / eV) = 6.78 ×10−3eV For λ = 0.5 nm, the wavelength is 30 times smaller, so that the energy is 900 times larger. Thus K =6.10 eV. 12. For a circular orbit of radius r, the circumference is 2πr. If n wavelengths λ are to fit into the orbit, we must have 2πr = nλ = nh/p. We therefore get the condition pr = nh / 2π = n